SME
Propulsion · Problem type
Two-stage rocket
Use this for “Stage 1 (300 s) takes a 120 t rocket to 50 t; stage 2 (340 s) takes 30 t to 10 t. What is the total Δv?”
Diagram
Δv = 2,577 m/s + 3,664 m/s
Δv = 6,241 m/s
This is an example. Type your own values to start a new problem.
Δv₁ (m/s)
=
Iₛₚ₁ (s)
×
g
×
ln(
m₀₁ (t)
÷
mf1 (t)
)
Δv₂ (m/s)
=
Iₛₚ₂ (s)
×
g
×
ln(
m₀₂ (t)
÷
mf2 (t)
)
Δv (m/s)
=
Δv₁ (m/s)
+
Δv₂ (m/s)
Formulas
Δv₁
=
Iₛₚ₁
×
9.81
×
ln(
m₀₁
mf1
)
2,576.5045
=
300
×
9.81
×
ln(
120
50
)
Δv₂
=
Iₛₚ₂
×
9.81
×
ln(
m₀₂
mf2
)
3,664.3114
=
340
×
9.81
×
ln(
30
10
)
Δv = Δv₁ + Δv₂
6,240.8159 = 2,576.5045 + 3,664.3114
Assumptions
•
Ideal Δv for each stage: no gravity or drag losses.
•
Stage 1’s empty structure drops at burnout; stage 2’s starting mass is what is left.
•
g = 9.81 m/s² defines specific impulse, so vₑ = Iₛₚ × g.
Step-by-step
WE KNOW
Iₛₚ₁ = 300 s, m₀₁ = 120 t, mf1 = 50 t, Iₛₚ₂ = 340 s, m₀₂ = 30 t, mf2 = 10 t
FIND
Δv of stage 1 (Δv₁), Δv of stage 2 (Δv₂), total Δv (Δv)
Step 1 · Find Δv of stage 1 (Δv₁)
Use Δv₁ = Iₛₚ₁ × 9.81 × ln(m₀₁ ÷ mf1)
Integrating
m
dv
=
−vₑ
dm
from
m₀
down
to
mf
gives
vₑ
ln(
m₀
mf
),
with
vₑ
=
Iₛₚ
×
g.
Δv₁
=
300
×
9.81
×
ln(
120
50
)
Δv₁ = 300 × 9.81 × ln(2.4)
Δv₁ = 300 × 9.81 × 0.8755
Δv₁ = 2,943 × 0.8755
Δv₁ = 2,576.504 m/s
Step 2 · Find Δv of stage 2 (Δv₂)
Use Δv₂ = Iₛₚ₂ × 9.81 × ln(m₀₂ ÷ mf2)
Integrating
m
dv
=
−vₑ
dm
from
m₀
down
to
mf
gives
vₑ
ln(
m₀
mf
),
with
vₑ
=
Iₛₚ
×
g.
Δv₂
=
340
×
9.81
×
ln(
30
10
)
Δv₂ = 340 × 9.81 × ln(3)
Δv₂ = 340 × 9.81 × 1.0986
Δv₂ = 3,335.4 × 1.0986
Δv₂ = 3,664.311 m/s
Step 3 · Find total Δv (Δv)
Use Δv = Δv₁ + Δv₂
Each stage adds its own Δv.
Δv = 2,576.504 + 3,664.311
Δv = 6,240.816 m/s
Check
Put the numbers back in. Both sides should match.
2,576.504
=
300
×
9.81
×
ln(
120
50
)
✓
3,664.311
=
340
×
9.81
×
ln(
30
10
)
✓
6,240.816 = 2,576.504 + 3,664.311 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.